Python sqlite3 库不能使用 URI，即使 sqlite3 版本应该能够

Question

I have the most recent sqlite3 installed on my machine:

$ sqlite3 --version
3.26.0 2018-12-01 12:34:55 bf8c1b2b7a5960c282e543b9c293686dccff272512d08865f4600fb58238b4f9

And, in python, the sqlite3 module is using this version of sqlite3:

$ python
Python 3.4.9 (default, Jan  5 2019, 18:35:56)
[GCC 5.5.0] on linux
Type "help", "copyright", "credits" or "license" for more information.
>>> import sqlite3 as sq
>>> sq.sqlite_version_info
(3, 26, 0)
>>> sq.version_info
(2, 6, 0)

However, I cannot open a database file using URIs, even though that feature has been present in sqlite since version 3.7:

>>> import sqlite3 as sq
>>> c = sq.connect('file://test', uri=True)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
sqlite3.NotSupportedError: URIs not supported

What is going on here? What have I done wrong?

Answer 1

OK, I figured out what was going on by reading the source code. When pyenv compiled my version of Python, the _sqlite module was compiled against the ridiculously old CentOS version of the "sqlite3.h" file. Because of this, the Python module didn't have the SQLITE_OPEN_URI macro defined, which causes it give a hard-coded "URIs not supported" Python exception.

To get around this, I had to set the following environment variable:

# This is to direct pyenv to the linuxbrew include and library directories, when building versions of Python
export PYTHON_CONFIGURE_OPTS="LD_RUN_PATH=/home/linuxbrew/.linuxbrew/lib/ LDFLAGS=-L/home/linuxbrew/.linuxbrew/lib/ CPPFLAGS=-I/home/linuxbrew/.linuxbrew/include/"